Aball is thrown with initial speed v0 up an inclined plane. the plane is inclined at an angle φ above the horizontal, and the ball’s initial velocity is at an angle θ above the plane. choose axes with x measured up the slope and y normal to the slope. you can neglect the z coordinate as the ball remains in the xy-plane (z = 0).

(a) write down newton’s second law using these axes and find the ball’s position as a function of time. (your solution only needs to be valid for times before the ball lands back on the inclined plane.)

(b) show that the ball lands on the plane a distance r = 2v0 2 sinθ cos(θ + φ) / (gcos2 φ) from its launch point.

(c) show that for given values of v0 and φ, the maximum possible range up the inclined plane is rmax = v0 2 /[g(1+ sinφ)].

Explanation:

a )  The plane is making angle φwith horizontal and ball is thrown with velocity v₀  at angle θ with the inclined plane . In this situation the only thing that should be taken care is that , instead of g , the component of g that is g cos φ will act on the object .

If t be the time after which ball reaches position y with respect to inclined plane

initial velocity with respect to inclined plane

u = v₀ sinθ vertically upwards with respect to inclined plane

Apply the formula

s = ut -1/2 gt²

y =  v₀ sinθ t - 1/2 gcosφ t²

b )

Displacement along the inclined plane R

Time of flight T

Use the formula

v = u - gt

0 = v₀ sinθ - gcosφ t

t = v₀ sinθ / gcosφ

Time of flight

T = 2t = 2 x v₀ sinθ / gcosφ

R = v₀ cosθ X T - 1/2  gsinφ x ( T)²

= 2v₀ cosθ x v₀ sinθ / gcosφ - 2  gsinφ x v₀² sin²θ / g²cos²φ

= 2v₀ ² sinθ (  cosθ / gcosφ - sinθsinφ /  gcos²φ )

= 2v₀ ² sinθ (cosθcosφ - sinθsinφ ) / gcos²φ

R = 2v₀ ²  sinθ cos(θ+ φ ) / gcos²φ

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

talk back to him about the sport and show him your expertise or leave the room